IBDP Maths SL 3.4 The circle radian measure of angles AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Explanation:
(a) Proving the equation:
1. Perimeter equation: \[ 2r + rθ = 10 \] \[ θ = \frac{10 – 2r}{r} \]
2. Area equation: \[ \frac{1}{2}r^2θ = 6.25 \] Substitute θ from perimeter: \[ \frac{1}{2}r^2\left(\frac{10 – 2r}{r}\right) = 6.25 \] Simplify: \[ \frac{10r – 2r^2}{2} = 6.25 \] \[ 5r – r^2 = 6.25 \] Rearrange: \[ 4r^2 – 20r + 25 = 0 \quad \text{(Proved)} \]
(b) Solving for r and θ:
1. Solve quadratic equation: \[ 4r^2 – 20r + 25 = 0 \] \[ (2r – 5)^2 = 0 \] \[ r = \frac{5}{2} \text{ cm} \]
2. Find θ using perimeter: \[ θ = \frac{10 – 2(\frac{5}{2})}{\frac{5}{2}} = 2 \text{ radians} \]
Question
A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.
The angle of the largest sector is twice the angle of the smallest sector.
Find the size of the angle of the smallest sector.
▶️Answer/Explanation
Markscheme
METHOD 1
If the areas are in arithmetic sequence, then so are the angles. (M1)
\( \Rightarrow {S_n} = \frac{n}{2}(a + l) \Rightarrow \frac{{12}}{2}(\theta + 2\theta ) = 18\theta \) M1A1
\( \Rightarrow 18\theta = 2\pi \) (A1)
\(\theta = \frac{\pi }{9}\) (accept \(20^\circ \)) A1
[5 marks]
METHOD 2
\({{\text{a}}_{12}} = 2{a_1}\) (M1)
\(\frac{{12}}{2}({a_1} + 2{a_1}) = \pi {r^2}\) M1A1
\(3{a_1} = \frac{{\pi {r^2}}}{6}\)
\(\frac{3}{2}{r^2}\theta = \frac{{\pi {r^2}}}{6}\) (A1)
\(\theta = \frac{{2\pi }}{{18}} = \frac{\pi }{9}\) (accept \(20^\circ \)) A1
[5 marks]
METHOD 3
Let smallest angle = a , common difference = d
\(a + 11d = 2a\) (M1)
\(a = 11d\) A1
\({S_n} = \frac{{12}}{2}(2a + 11d) = 2\pi \) M1
\(6(2a + a) = 2\pi \) (A1)
\(18a = 2\pi \)
\(a = \frac{\pi }{9}\) (accept \(20^\circ \)) A1
[5 marks]
▶️ Answer/Explanation
a. Using the area formula for triangle AOP with angle \( \theta \): \[ \text{Area} = \frac{1}{2}r^2 \sin \theta \]
b. For triangle POT, first find TP using tangent: \[ \text{TP} = r \tan \theta \] Then area: \[ \text{Area} = \frac{1}{2}r \times r \tan \theta = \frac{1}{2}r^2 \tan \theta \]
c. Comparing areas: \[ \text{Area of triangle AOP} < \text{Area of sector} < \text{Area of triangle POT} \] \[ \frac{1}{2}r^2 \sin \theta < \frac{1}{2}r^2 \theta < \frac{1}{2}r^2 \tan \theta \] Dividing by \( \frac{1}{2}r^2 \): \[ \sin \theta < \theta < \tan \theta \]
Markscheme:
a. \( \frac{1}{2}r^2 \sin \theta \) A1 [1 mark]
b. • Correct TP (M1)
• \( \frac{1}{2}r^2 \tan \theta \) A1 [2 marks]
c. • Correct sector area A1
• Correct inequality chain R1
• Final proof AG [2 marks]
▶️ Answer/Explanation
Markscheme:
Area of triangle \( = \frac{1}{2}{(2x)^2}\sin \frac{\pi }{3}\) (M1)
\( = {x^2}\sqrt 3 \) A1
Note: A \(0.5 \times {\text{base}} \times {\text{height}}\) calculation is acceptable.
Area of sector \({\text{ = }}\frac{\theta }{2}{r^2} = \frac{\pi }{6}{r^2}\) (M1)A1
Area of triangle is twice the area of the sector
\( \Rightarrow 2\left( {\frac{\pi }{6}{r^2}} \right) = {x^2}\sqrt 3 \) M1
\( \Rightarrow r = x\sqrt {\frac{{3\sqrt 3 }}{\pi }} \,\,\,\,\,\)or equivalent A1
[6 marks]